I was wondering about one step in the proof of surjectivity of $\Theta$ constructed for Theorem 3.4.3 in Weibel's "An introduction to homological Algebra".
For an extension $\xi:0\to B\to X\to A\to0$ of $A$ by $B$, he associates the element $x=\Theta(\xi)\in \operatorname{Ext}^1(A,B) $, by applying $\operatorname{Ext}^*(A,-)$ to form the long exact sequence
$$\cdots\to\operatorname{Hom}(A,X)\to \operatorname{Hom}(A,A)\xrightarrow{\partial}\operatorname{Ext}^1(A,B)\to\cdots, $$
and setting $x=\partial(\operatorname{id}_A)$.
He then shows that $\Theta$ gives a well defined map from the set of equivalence classes of extensions of $A$ by $B$ to $\operatorname{Ext}^1(A,B)$.
In proving surjectivity of $\Theta$, he considers an exact sequence $0\to M\to P \to A\to 0$ with $P$ projective. Applying $\operatorname{Ext}^*(-,B)$
gives $\operatorname{Hom}(M,B)\xrightarrow{\partial}\operatorname{Ext}^1(A,B)\to 0$, so for $x\in \operatorname{Ext}^1(A,B)$ he picks $\beta:M\to B$ with $x=\partial(\beta)$. Then he constructs a diagram
\begin{array}{ccccccccc}
0 & \xrightarrow{} & M & \xrightarrow{} & P & \xrightarrow{} & A & \xrightarrow{} & 0\\
& & \downarrow & & \downarrow & & \parallel & & \\
0 & \xrightarrow{} & B & \xrightarrow{} & X & \xrightarrow{} & A & \xrightarrow{} & 0
\end{array}
with the map from $M$ to $B$ given by $\beta$ and $X$ is the pushout of $B\leftarrow M\to P$. One shows that the lower row is exact (no problem).
Then he claims that the extension given by the lower row maps to $x$ under $\Theta$. How does this follow? He states that one uses the naturality of $\partial$, so one has to apply $\operatorname{Ext}$ in some way.
Edit:Taking the Ext long exact sequence doesn't solve the problem immediately, see the comments below.
Best Answer
The proof as intended by Weibel doesn't seem to work, since it requires one to solve the related question asked here Are those two ways to relate Extensions to Ext equivalent?. However, by using the dual version of his proof we can avoid this issue:
Pick an exact sequence $0\to B \to I\xrightarrow{\pi} N\to 0$, where now $I$ is an injective object. Then apply $Ext(A,-)$ to obtain en exact sequence $$ ... \to Hom(A,N) \xrightarrow{\partial} Ext(A,B) \to 0,$$ and pick $\gamma \in Hom(A,N)$ with $\partial(\gamma)=x$. Now we let $X$ be the pullback of $A\xrightarrow{\gamma} N \xleftarrow{\pi}I$. This fits into a commutative diagram with exact rows:
\begin{array}{ccccccccc} 0 & \xrightarrow{} & B & \xrightarrow{} & X & \xrightarrow{} & A & \xrightarrow{} & 0\\ & & \parallel & & \downarrow & & \downarrow & & \\ 0 & \xrightarrow{} & B & \xrightarrow{} & I & \xrightarrow{} & N & \xrightarrow{} & 0. \end{array} The upper row is now an extension $\xi$ for which one directly sees that $\Theta(\xi)=x$: Applying $Ext(A,-)$ again gives a long ladder diagram, from which we consider the square \begin{array}{ccc} Hom(A,A) & \xrightarrow{\partial'} & Ext^1(A,B) \\ \downarrow& & \parallel \\ Hom(A,N) &\xrightarrow{\partial} & Ext^1(A,B) \end{array} The $\partial$ here is the same as above, and by the definition in Weibel we have $\Theta(\xi) =\partial'(id_A)$. Finally, the left vertical arrow is composition with $\gamma$ by definition of the $Hom$ functor. So $$\Theta(\xi) =\partial'(id_A)=\partial(\gamma\circ id_A) =x.$$